
\section{Still refreshing ...}

\subsection*{Problem 1}

Consider the possible cases:
\begin{itemize}
  \item[case] "H", probability: $\frac{1}{2}$
  \item[case] "TH", probability: $\frac{1}{4}$
  \item[case] "TTH", probability: $\frac{1}{8}$
  \item[case] "TTTH", probability: $\frac{1}{16}$
\end{itemize}
and so on. For the expected number of heads, this yields the infinite series:
\begin{equation}
  E[H] = \sum_x x \cdot p(x) = \sum_{n=0}^\infty 1 \cdot \frac{1}{2^n} = 
         \sum_{n=0}^\infty \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 
         \frac{2}{2} = 1,
\end{equation}
where we used the formula for infinite geometric series.

In the same way, for the expected number of tails we get:
\begin{equation}
  E[T] = \sum_x x \cdot p(x) = \sum_{n=0}^\infty n \cdot \frac{1}{2^{n+1}} = 1,
\end{equation}
where we used Wolfram Mathematica to evaluate the infinite sum.

So, $E[H] = 1, E[T] = 1$.





\subsection*{Problem 2}
Probability of using urn $u$: $p(u)=\frac{1}{11}$

Probability of extracting $n_{B}$ black balls and $n_{W}$ from urn
$u$: 
\[
p(n_{B},n_{W}|u)=\left(\frac{u}{10}\right)^{n_{B}}\left(\frac{10-u}{10}\right)^{n_{W}}\binom{10}{n_{B}}
\]

Probability of using urn $u$ given $N=10$ extractions, where $3$
of the drawn balls were black ($n_{B}=3$, $n_{W}=N-n_{B}=7$):
\[
p(u|n_{B},n_{W})=\frac{\left(\frac{u}{10}\right)^{n_{B}}\left(\frac{10-u}{10}\right)^{n_{W}}\binom{10}{n_{B}}p(u)}{\sum_{i=0}^{10}\left(\frac{i}{10}\right)^{n_{B}}\left(\frac{10-i}{10}\right)^{n_{W}}\binom{10}{n_{B}}p(i)}=\frac{\left(\frac{u}{10}\right)^{n_{B}}\left(\frac{10-u}{10}\right)^{n_{W}}}{\sum_{i=0}^{10}\left(\frac{i}{10}\right)^{n_{B}}\left(\frac{10-i}{10}\right)^{n_{W}}}
\]
\[
p(u|3,7)=\frac{\left(\frac{u}{10}\right)^{3}\left(\frac{10-u}{10}\right)^{7}}{\sum_{i=0}^{10}\left(\frac{i}{10}\right)^{3}\left(\frac{10-i}{10}\right)^{7}}
\]

The probability of alice drawing another black ball is:
\[
p(1_B | u) = \frac{u}{10} p(u | n_B , n_W)
\]
